Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int inf = 1e9;
const int MOD = 1e9 + 7;
const int N = 1e5 + 10;
int b[N], lv[N], ans[N], a[N], p[N][20], in[N], out[N], n, m, k;
int c[N][4], need[N], L[N], R[N], s[N];
vector<int> g[N], v[N];
queue<int> q[N];
int get(int x) {
  int ans = 0;
  while (x) {
    ans += b[x];
    x -= x & -x;
  }
  return ans;
}
void add(int x, int v) {
  while (x <= n) {
    b[x] += v;
    x += x & -x;
  }
}
void DFS(int x) {
  static int top = 0;
  in[x] = ++top;
  lv[x] = lv[p[x][0]] + 1;
  for (auto y : v[x])
    if (y != p[x][0]) {
      p[y][0] = x;
      DFS(y);
    }
  out[x] = top;
}
int LCA(int x, int y) {
  if (lv[x] < lv[y]) swap(x, y);
  for (int i = 0, v = lv[x] - lv[y]; v; i++, v >>= 1)
    if (v & 1) x = p[x][i];
  if (x == y) return x;
  for (int i = int(log2(lv[x])); i >= 0; i--)
    if (p[x][i] != p[y][i]) x = p[x][i], y = p[y][i];
  return p[x][0];
}
void prepare() {
  scanf("%d%d%d", &n, &m, &k);
  for (auto i = 1; i <= n - 1; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    v[x].push_back(y);
    v[y].push_back(x);
  }
  DFS(1);
  for (auto lv = 1; lv <= 18; lv++)
    for (auto i = 1; i <= n; i++) p[i][lv] = p[p[i][lv - 1]][lv - 1];
  for (auto i = 1; i <= m; i++) scanf("%d", a + i);
  for (auto i = 1; i <= k; i++) {
    scanf("%d%d%d", c[i], c[i] + 1, need + i);
    c[i][2] = LCA(c[i][0], c[i][1]);
    c[i][3] = p[c[i][2]][0];
  }
}
void sweep(int phase) {
  memset(b, 0, sizeof(b));
  for (auto i = 1; i <= m; i++) {
    add(in[a[i]], 1);
    add(out[a[i]] + 1, -1);
    while (!q[i].empty()) {
      int idx = q[i].front();
      q[i].pop();
      int tmp[4];
      for (auto i = 0; i <= 3; i++) tmp[i] = get(in[c[idx][i]]);
      if (tmp[0] + tmp[1] - tmp[2] - tmp[3] >= phase)
        R[idx] = i - 1;
      else
        L[idx] = i + 1;
    }
  }
}
void solve() {
  for (auto phase = 1; phase <= 10; phase++) {
    int top = 0;
    for (auto i = 1; i <= k; i++)
      if (!ans[i]) {
        s[++top] = i;
        L[i] = 1, R[i] = m;
      }
    while (top) {
      for (auto j = top; j >= 1; j--) {
        int i = s[j];
        if (L[i] <= R[i]) {
          int mid = (L[i] + R[i]) >> 1;
          q[mid].push(i);
        } else {
          if (L[i] <= m) g[i].push_back(L[i]);
          if (L[i] > m || need[i] == phase) ans[i] = 1;
          swap(s[j], s[top--]);
        }
      }
      sweep(phase);
    }
  }
  for (auto i = 1; i <= k; i++) {
    printf("%d ", int(g[i].size()));
    for (auto x : g[i]) printf("%d ", x);
    printf("\n");
  }
}
int main() {
  prepare();
  solve();
}